For homework i have to compute the homology groups of the quotient space obtained by gluing the boundary of the Möbius band to the circle $x^2 + y^2 =1$ of $S^2$. But I find a bit difficult to understand what is the space resulting from this quotient. First I thought it was a real proyective plane but then i realized it's not truth.

After the hits of Artic char, I can do the following

Let $X$ be the quotient space that is obtained by gluing the boundary of the Möbius strip to the circle of .

Let $U= X \backslash${equator of the Möbius strip} and $V= X \backslash${north pole and south pole of }. Note that $U \cup V =X$ and $U \cap V$ is homotopic to $S^1$. It is easy to see that $V$ is homotopic to the Möbius strip, which in turn is homotopic to $S^1$, while $U$ is homotopic to $S^2$. In this way we have the following Mayer-Vietoris sequence,

$H_2 (S^1) \longrightarrow H_2 (S^2) \oplus H_2 (S^1) \longrightarrow H_2 (X) \longrightarrow H_1 (S^1) \longrightarrow H_1 (S^2) \oplus H_2 (S^1) \longrightarrow H_1 (X)$

With the previous chain we have the following chain,

$0 \longrightarrow \mathbb{Z} \longrightarrow H_2(X) \longrightarrow \mathbb{Z} \longrightarrow \mathbb{Z} \longrightarrow H_1(X) \longrightarrow \mathbb{Z}$

Since we have an exact sequence, it follows that $\ker\delta_ {i} = \mathrm{img} ~ \delta_{i+1}$

but i don't know how to continue. Could you help me a little?

  • 1
    $\begingroup$ Can you use mayer vietoris? $\endgroup$ Jun 4, 2022 at 23:32
  • $\begingroup$ Yes!, actually I was using MV, but I can't understand the resulting space from this quotient, for that reason I can't decompose that space as the union of two subspaces. Do you have any hint? $\endgroup$
    – IMMA
    Jun 4, 2022 at 23:36
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    $\begingroup$ Let $M$ be your space, and let $U = M \setminus\{\text{equator of the Mobious band}\}$ and $V = M \setminus \{\text{ North and south pole of }S^2\}$. Note that $U\cup V = M$ and $U\cap V$ is homotopic to $S^1$. $\endgroup$ Jun 4, 2022 at 23:41
  • $\begingroup$ Ok. I understand. But know I think it is more complex than i thought because the $H_n (U)$ and $H_n(V)$ are not easy to compute, in order to use Mayer-Vietoris sequence. The $H_n ( U \cup V)$ its very easy. Maybe there is someting I don't see yet. $\endgroup$
    – IMMA
    Jun 4, 2022 at 23:54
  • $\begingroup$ All are quite easy to compute.... for example $U$ is homotopic to $S^2$, while $V$ is homotopic to the mobius band (which is homotopic to $S^1$). $\endgroup$ Jun 5, 2022 at 0:00

1 Answer 1


This space is homeomorphic to $\mathbb {RP}^2$ with an extra $2$ cell attached in the same way as the usual one. So the chain complex is given by $\mathbb Z^2\to \mathbb Z \to\mathbb Z$ with the first map sending $(x,y)\mapsto 2x+2y$ and the second being the $0$ map. The kernel of the first map is $\mathbb Z$ generated by $(1,-1)$ and its image is $2\mathbb Z$. So we get $H_2=\mathbb Z$, $H_1=\mathbb Z/2\mathbb Z$.

Another way to see this is to notice that one can contract the attaching circle of the mobius strip to the sphere to a point, giving that the space is homotopy equivalent to $S^2\vee \mathbb {RP}^2$.

Edit: In order to apply MV to the original space, you need to understand what your $\mathbb Z\to\mathbb Z$ map is. This comes from the map on homology which includes the boundary of the mobius strip into the mobius strip. The boundary circle wraps twice around the core circle, so this map is multiplication by $2$. Thus it has kernel $0$, which means that the previous $\mathbb Z\to H_2(X)$ is an isomorphism. The map from $H_1(X)$ is $0$ (which you can see by considering reduced homology) so we have an exact sequence $0\to \mathbb Z\overset{\times 2}{\to} \mathbb Z\to H_1(X)\to 0$, implying $H_1=\mathbb Z/2\mathbb Z$.

  • $\begingroup$ Thank you for your answer. I have a question who are U and V in this case. And, do you think that my idea is wrong? :( $\endgroup$
    – IMMA
    Jun 5, 2022 at 3:16
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    $\begingroup$ I did not use Mayer Vietoris in my argument. In the simplest form I just argued that the space is homotopy equivalent to the wedge of a sphere and the projective plane. From there you can use the fact that the homology of a wedge is the direct sum of the homologies of the individual spaces. (Which can be proven using MV.) I'll add a bit on how to use MV in the original. $\endgroup$ Jun 5, 2022 at 3:39
  • $\begingroup$ Thank you so much! I understand perfectly. $\endgroup$
    – IMMA
    Jun 5, 2022 at 4:29

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